Statistical Mechanics. Hagai Meirovitch

Transcription

1 Statistical Mechanics Hagai Meirovitch 1

2 IRODUCIO Classical thermodynamics is a general methodology that enables one to derive relations among macroscopic properties such as volume (V), pressure (P), temperature (), energy (E), entropy (S), Helmholtz free energy (A=E-S) etc. without the need to consider the atomic properties of the system: e.g., the equation of state is PV=R. Some of other relations: V A P, = V A E, ) ( = V A S, =

3 In statistical mechanics, on the other hand, the macroscopic thermodynamic behavior is obtained from the specific microscopic description of a system (atomic interactions, atomic masses, etc.). herefore, statistical mechanics also provides information that is beyond the reach of classical thermodynamics - what is the native structure of a protein? It is impossible to know the exact dynamic behavior of a large system (i.e., the atomic coordinates at time t). Appearance of a 3D configuration is only known with a certain probability. hus, statistical mechanics is a probabilistic theory, and computational methods, such as Monte Carlo and molecular dynamics are based on these probabilistic properties. 3

4 Program (a) We start by summarizing some properties of basic physics of mechanical systems. (b) A very short discussion on classical thermodynamics. (c) Derivation of statistical mechanics by demanding compatibility with the theory of classical thermodynamics. (d) Statistical mechanics as a probabilistic theory. (e) Examples. 4

5 Macroscopic Mechanical Systems are Deterministic Refreshing some basic physics Examples of forces F (F= F ): 1) Stretching a spring by a distance x: F=-fx, Hook s Law f - spring constant. ) Gravitation force: F=kMm/r - m and M masses with distance r; k - constant. On earth (R,M large), g=km/r F=mg 3) Coulomb law: F=kq 1 q /r q 1, q charges. 5

6 ewton s second law: F=ma - a acceleration Mechanical work W: if a constant force is applied along distance d, W=Fd (F= F ). More general, W=! F. dx. Potential energy: If mass m is raised to height, h negative work is done, W = mgh and the mass gains potential energy,e p = -W = +mgh - the ability to do mechanical work: when m falls dawn, E p is converted into: kinetic energy, E k = mv /, where v /=gh (at floor). A spring stretched by d: E p = -W = f! xdx = fd / In a closed system the total energy, E t = E p + E k is constant but E p /E k can change; e.g., oscillation of a mass hung on a spring 6 and distorted from its equilibrium position.

7 he dynamical state of a mechanical macroscopic system in principle is deterministic in the sense that if the forces are known, and the positions and velocities of the masses at time t=0 are known as well, their values at time t can in principle be determined by solving ewton s equations, F=ma. Simple examples: harmonic oscillator (a spring), a trajectory of a projectile, movement of a spaceship, etc. In some cases the solution might be difficult requiring strong computers. 7

8 Stability - a system of interacting masses (by some forces) tends to arrange itself in the lowest potential energy structure (which might be degenerate) also called the ground state. he system will stay in the ground state if the kinetic energy is very small - this situation defines maximum order. he larger the kinetic energy the larger is the disorder -in the sense that at each moment a different arrangement of the masses will occur (no ground state any more). Still, in principle, the trajectories of the masses can be calculated. wo argon atoms at rest positioned at the lowest energy distance ε interacting through Lennard-Jones potential. Microscopic system. σ φ(r) = 4ε r 1 σ r 6 ε 8

9 Classical hermodynamics Unlike mechanical systems, thermodynamic systems also include transition of heat. Definition of the (dynamic) state by coordinates and velocities of the molecules is impossible. A state is defined by several macroscopic variables only: volume V, temperature, pressure P that are not independent - connected by an equation of state f(p,v,) = 0 Equilibrium external conditions have not changed for a long time and all relaxation processes have occurred. Reversible transformation from states A very slow, actually composed of many reversible intermediate states. Very general theory because no microscopic (atomic) information is involved. 9

10 he first law of thermodynamics he first law is the conservation of energy for thermodynamic systems. It states that the variation in energy during any transformation (from state a to b) is equal to the amount of energy that the system receives from the surrounding, and it depends on states a and b and not on the intermediate process. he law implies the existence of internal energy E which is a state function defined up to an additive constant and is an extensive parameter (proportional to the size of the system), E - E 1 = ΔE = Q W Q- heat received from the surrounding; W - mechanical work done by the system. is of surrounding. 10

11 he second and third laws of thermodynamics he law is based on the realization that while mechanical work can be transferred completely to heat, there is a limitation in transferring heat to mechanical work. From the microscopic point of view - molecular motion is not ordered and the kinetic energy cannot be fully harnessed. he law was introduced initially in terms of engines, perpentum mobile, Carnot cycle, by Kelvin, Clausius and others. will represent a postulate based on entropy. A given thermal equilibrium state a is chosen as standard state. he entropy S(b) of the system in state b is = b d ' Q Integration over any reversible path a b S ( b ) a is of surrounding, d Q absorbed by system S(=0)=0 (third law). Entropy is an extensive state function; he second law: b d' Q S( b) S( a) equality - for a reversible path 11 a

12 Helmholtz free energy F (A), independent variables and V F = E-S Helmholtz free energy. F = F(,V,) F V = S F V = P Maxwell relation S V P = V Suppose a system that cannot exchange energy in the form of work (W) with the environment. is defined by environment. In this case 0 = W F(a) F(b) F(b) F(a) the free energy cannot increase during a transformation. OR If the free energy is a minimum the system is in a state of stable equilibrium. ecause any transformation would produce an increase in F -in contradiction to the above inequality. 1

13 hermodynamic systems at equilibrium and equilibrium statistical mechanics A typical system is described below; examined from a microscopic point of view non-rigorous treatment R R C C A system C of molecules in a constant volume V is in thermal contact with a large reservoir (R) (also called heat bath) with a well defined temperature R. At equilibrium (after a long time) energy is still exchanged between R and C but the average kinetic energy of a molecule of C is constant and equal to that in R leading to c = R. 13

14 However, in contrast to a macroscopic mechanical system, there is no way to know the trajectories of the particles that are changed constantly due to collisions and the energy exchange between C&R, i.e. the random knocks of the R atoms on the interface. However, the averaging E k & E p along very long trajectory leads to definite <E k > and <E p >, and thus for a snapshot of configuration of particles ~<E k > and ~<E p >. Relating kinetic energy to temperature: At high, E k is high - dominant, E p is high strong random knocks and collisions, large uncertainty related to trajectories large disorder. At low, E k is low, the effect of the molecular forces significant - the system arranges itself in a low potential energy state low disorder. herefore, a thermodynamic system at equilibrium cannot be characterized by the positions & velocities of its 10 3 particles but only by the average values of several macroscopic parameters such as P,, E (internal energy) and entropy, S. 14

15 For example, in measuring the thermometer feels the average effect of many molecular states (x,v ) of the tested system and for a long measurement all the microscopic states (x,v ) are realized and affect the result of. Hence to obtain the average values of macroscopic parameters from microscopic considerations a probability density P(x,v ) should be assigned to each system state (x,v ) where (x,v ) = (x 1,y 1,z 1,x,y,z,.x,y,z,v x1,v y1 v z1. v x,v y v z ) thus assuming that all states contribute. he space based on (x,v ) is called phase space (with P a probability space, or ensemble) 15

16 hen a macroscopic parameter M is a statistical average, <M> =! P (x,v ) M(x,v ) d(x v ). he entropy for a discrete and continuous system is (k is the oltzmann constant), S = <S> = -k Σ P i ln P i and S = -k! P(x v ) lnp(x,v ) d(x v )+ + ln{const.[dimension (x v )]} otice, P is a probability density with dimension 1/ (x v ). he constant is added to make S independent of (x v ). 16

17 he problem how to determine P. he correct P should lead to the usual thermodynamic properties. In thermodynamics an,v, system is described by the Helmholtz free energy A, A(,V,)=E S, which from the second law of thermodynamics should be minimum for a given set of constraints. We shall determine P by minimizing the statistical free energy with respect to P. 17

18 A can be expressed as the average A = <A(P)>=! P(X)[E(X)+k lnp(x)]dx+const. X=(x,v ). We take the derivative of A with respect to P and equate to zero; the const. is omitted. A =! [E(X)+k lnp(x) + k P(X) /P(X)]dX=0 E(X)+k [lnp(x) + 1]=0 lnp(x) = - [E(X) + k ]/k = = - [E(X)/k ] +1 P(X) =constÿexp[-e(x)/k ] he normalization is Q=! exp[-e(x)/k ]dx 18

19 i.e., P (X)=exp[-E(X)/k ]/Q P the oltzmann probability (density). Q the canonical partition function. he intgrand defining A (p. 18) is E(X)+k lnp (X) Substituting P and taking the ln gives, E(X)+k [- E(X)/k lnq]= -k lnq -k lnq is constant for any X and can be taken out of the integral of A. hus, A=<E>-<S>= - k lnq 19

20 In summary. Assuming the existence of temperature and that the entropy function is S = -k! P(x v ) lnp(x,v ) d(x v ) We have defined P by connecting with thermodynamics. he relation A= - k lnq is very important. It enables calculating A by Q that is based on the microscopic details of the system. In classical thermodynamics all quantities are obtained as derivatives of A. Hence, with statistical mechanics all these quantities can be obtained by taking derivatives of lnq. 0

21 Also, the probabilistic character of statistical mechanics enables calculating averages even without calculating Q. hese averages can be even geometrical properties that are beyond the reach of thermodynamics, such as the end-to-end distance of a polymer, the radius of gyration of a protein and many other geometrical or dynamic properties. his is in particular useful in computer simulation. he ensemble of configurations defined by constant V, and and their oltzmann probabilities is called the canonical ensemble. 1

22 Calculation of the partition function Q R Systems at equilibrium particles with velocities v and coordinates x are moving in a container in contact with a reservoir of temperature R =. We have seen that the Helmholtz free energy, A is A=E-S= - k lnq where Q=! exp[-e(x,v )/k ]d(x v ) )

23 E(x,v )= E k (v ) + E p (x ) E(x,v ) is the Hamiltonian of the system (for E p (x ) see p. 8). If the forces do not depend on the velocities (most cases) E k is independent of E p and the integrations can be separated. Also, the integrations over the velocities of different particles are independent. Moreover, the integrations over the components v x,v y, and v z are independent. herefore, we treat first the integration over v x (denoted v) of a single particle. o recall: the linear momentum vector p=mv; therefore, for one component: E k = mv / = p /m 3

24 A useful integral (from table): exp( ax ) dx= 0 1 π a herefore, for calculating Q(p) our integral is because p exp( ) dp mk 1 a = mk = πmk 4

25 he integral over the 3 components of the momentum (velocities) is the following product: 3 ( πmk ) Q is (h is the Planck constant see Hill p.74; W=V ), Q = ( πmk ) 3! h 3 exp( Ω E(x k ) ) d x he problem is to calculate the configurational integral 5

26 E(x ) exp( ) dx 1 dy 1 dz 1 dx dy dz... Ω k he division by factorial,!=1μ μ 3 μ is required from quantum mechanical considerations because the particles are independent and indistinguishable (however, introduced by Gibbs!, Hill p.6). hus, each configuration of the particles in positions x 1, x, x 3,.,x can be obtained! times by exchanging the particles in these positions For example, for 3 particles there are 3!=6 possible permutations. sites Particles 1,,

27 Stirling approximate formula: ln!º ln (/e) he Helmholtz free energy A(,V,) is: 3/ πmk k ln{ h e } k ln exp Ω E(x ) dx k he velocities (momenta) part is completely solved; it contains m and h - beyond classical thermodynamics! he problem of statistical mechanics is thus to solve the integral. For an ideal gas, E = 0 (no interactions) hence Û=V trivial!! A(, V, ) = k πmk ln{ h 3/ ev } 7

28 hermodynamic derivatives of A of an ideal gas (properties ~ or ~V are called extensive): Pressure: Intensive variable ~/V Internal energy: Extensive variable ~ A P= = k / V PV= V E =, A ( ) V, = 3 k k E is the average kinetic energy (see later) ~, independent of V! Each degree of freedom contributes (1/)k. If the forces (interactions) do not depend on the velocities, is determined 8 by the kinetic energy only (see previous note).

29 he specific heat C V is independent of and V. A = C V he entropy is: S V, = = E E A V, = k = 3 k πmk ln h 3/ Ve In thermodynamics one obtains for an ideal gas: S=C v ln+rlnv+a. S contains microscopic parameters (e.g., m) not known in thermodynamics. S increases with - extensive variable. V/ intensive! S is not defined at =0 should be 0 according the third law of thermodynamics; the ideal gas picture holds only for high. S ~ ln and E~, both increase with, but A=E-S. 9 5/

30 In the case of a real gas E(x ) 0 and the problem is to calculate the configurational partition function denoted Z, where the momentum part is ignored, Z=! exp[- E(x )/k ]dx otice: While Q is dimensionless, Z has the dimension of x. Also, Z=! n(e)exp[- E/k ] de n(e) the density of states around E; n(e)de the volume in configurational space with energy between E and E+ de. For a discrete system [n(e i ) is the degeneracy of E] Z = Ê exp [- E i /k ] = Ê n(e i ) exp [- E i /k ] 30

31 he contribution of Z to the thermodynamic functions is obtained from derivatives of the configurational Helmholtz free energy, F F = -k ln Z Calculating Z for realistic models of fluids, polymers, proteins, etc. by analytical techniques is unfeasible. Powerful numerical methods such as Monte Carlo and molecular dynamics are very successful. A simple example: classical harmonic oscillators hey are at equilibrium with a heat bath of temperature -a good model for a crystal at high temperature, where each atom feels the forces exerted by its neighbor atoms and can approximately be treated as an independent oscillator that does not interact with the neighbor ones. 31

32 herefore, Q =q, where q is the partition function of a single oscillator. Moreover, the components (x, y, z) are independent as well; therefore, one can calculate q x and obtain Q =q x 3. he energy of a macroscopic oscillator (e.g., a mass hung on a spring) is determined by its amplitude (the stretching distance) and the position x(t) of the mass is known exactly as a function of t. he amplitude of a microscopic oscillator is caused by the energy provided by the heat bath. his energy changes all the time and the amplitude changes as well but has an average value that increases as increases. Unlike a macroscopic mechanical oscillator, x(t) is unknown; we only know P (x). 3

33 Oscillators in thermal equilibrium with a heat bath of temperature. A snapshot of their amplitudes at time t. he average amplitude is well defined (or the average amplitude of one oscillator as a function of time)- 33 canonical ensemble picture.

34 he kinetic and potential energy (Hamiltonian) of an oscillator are p /m + fx / f is the force constant and the partition function is, q=q k q p where q k was calculated before; n is the frequency of the oscillator. q k mk = π h k qp = π f q= π mk f h k = hν 1 ν = π f m 34

35 A E = = k E k = E p = 1/k fl C V = k he average energy (see later) of one component (e.g., x direction) of an oscillator is twice as that of an ideal gas effect of interaction. For, 3D oscillators, E= 3k - extensive (~) and C V = 3k - Dulong-Petit Law. he entropy is also extensive, S = E/- A/=3k (1+ln [k /hn] ) E and S increase with. In mechanics the total energy of an oscillator is constant, fd / where d is the amplitude of the motion and at time t the position of the mass is known exactly. 35

36 In statistical mechanics a classical oscillator changes its positions due to the random energy delivered by the heat bath. he positions of the mass x are only known with their Gaussian oltzmann probability; P(x) = exp[-fx /k ]/q p In principle x can be very large, but for a given there is an average amplitude & in practice x fluctuates around this value. When increases the energy increases meaning that the average amplitude increases and the position of the mass is less defined due to larger fluctuations (in x ); therefore, the entropy is enhanced. otice: A classical oscillator is a valid system only at high. At low one has to use the quantum mechanical oscillator. 36

37 hus far we have obtained the macroscopic thermodynamic quantities from a microscopic picture by calculating the partition function Q and taking (thermodynamic) derivatives of the free energy kln Q. We have discussed two simple examples, ideal gas and classical oscillators. We shall discuss next the probabilistic significance of statistical mechanics. 37

38 he probabilistic character of statistical mechanics hus far, the thermodynamic properties were obtained from the relation between the free energy, A and the partition function Q, A=-k lnq using known thermodynamics derivatives. However, the theory is based on the assumption that each configuration in phase space has a certain probability (or probability density) to occur - the oltzmann probability, P (X)=exp[-E(X)/k ]/Q where X ª x v is a 6 vector of coordinates and velocities herefore, any thermodynamic property such as the energy is an expectation value defined with P (X). 38

39 39 For example, the statistical average (denoted by <>) of the energy is: <E>=! P (x,v ) E(x,v ) d (x v ), where E(x,v ) is a random variable. <E> is equal to E calculated by a thermodynamic derivative of the free energy, A. For an ideal gas (pp.-5) <E> is obtained by a Gaussian k V mk d d mk m E V mk mk Q h mk P V h mk Q V 3 ) ( ] [ exp ) ( ] [ exp! ]/ [ exp! ) ( 3 3 IG IG = π >= < π = = π = x p p p p p

40 40 his is the same result obtained from thermodynamic derivatives (p. 8). For two degrees of freedom the integral is: k mk m mk mk V mk dx dx dp dp mk p p m p p 3 ) ( )] ( ) ( [ ) ( ] [ exp ) ( 3/ = π π π = = π + +

41 x exp [ ax ] dx = a π 3 / 41

42 he entropy can also be expressed as a statistical average. For a discrete system, S =<S> = -k Σ i P i ln P i If the system populates a single state k, P k =1 and S=0 fl there is no uncertainty. his never occurs at a finite temperature. It occurs for a quantum system at =0 K. On the other hand, if all states have the same probability, P i =1/W, where W is the total number of states, the uncertainty about the state of the system is maximal (any state can be populated equally) and the entropy is maximal as well, S= -k ln W his occurs at very high temperatures where the kinetic energy is large and the majority of the system s configurations can be visited with the same random probability. 4

43 It has already been pointed out that the velocities (momenta) part of the partition function is completely solved (pp. 4-6). On the other hand, unlike an ideal gas, in practical cases the potential energy, E(x ) 0 and the problem of statistical mechanics is to calculate the configurational partition function denoted Z, where the momentum part is ignored (see p. 30) Z=! exp[- E(x )/k ]dx where Z has the dimension of x. Also, Z=! n(e)exp[- E/k ] de n(e) the density of states; n(e)de the volume in configurational space with energy between E and E+ de. For a discrete system n(e i ) is the degeneracy and Z is Z=Êi exp [- E i /k ] = Ê E n(e) exp [- E/k ] 43

44 he thermodynamic functions can be obtained from derivatives of the configurational Helmholtz free energy, F F=-k ln Z. From now on we ignore the velocities part and mainly treat Z. Probability space picture: he configurational space is viewed as a 3 dimensional sample space W, where to each point x (random variable) corresponds the oltzmann probability density, P (x )=exp-[e(x )]/Z where P (x )dx is the probability to find the system between x and x + dx. he potential energy E(x ) defined for x is also a random variable with an expectation value <E> 44

45 he power of the probabilistic approach is that it enables calculating not only macroscopic thermodynamic properties such as the average energy, pressure etc. of the whole system, but also microscopic quantities, such as the average end-toend distance of a polymer. his approach is extremely useful in computer simulation, where every part of the system can be treated, hence almost any microscopic average can be calculated (distances between the atoms of a protein, its radius of gyration, etc.). he entropy can also be viewed as an expectation value, where ln P (x ) is a random variable, S = <S> = -k! W P (x )ln P (x )dx 45

46 Likewise, the free energy F can formally be expressed as an average of the random variable, E(x )+ k ln P (x ), F = - k lnz =! W P (x )[E(x )+k ln P (x )] dx Fluctuations (variances) he variance (fluctuation) of the energy is: s (E) =! W P (x )[E(x )-<E>] dx = = <E(x ) > - <E(x ) > otice that the expectation value is denoted by <> and E is the energy. It can be shown (Hill p. 35) that the specific heat at constant volume is: C v = (de/d) V = s (E) /k 46

47 In regular conditions C v = s (E) /k is an extensive variable, i.e., it increases ~ as the number of particles increases; therefore, s(e) ~ 1/ and the relative fluctuation of E decreases with increasing, σ( E) 1 ~ = < E > hus, in macroscopic systems ( ~10 3 ) the fluctuation (i.e., the standard deviation) of E can be ignored because it is ~10 11 times smaller than the energy itself fl these fluctuations are not observed in macroscopic objects. Like Z,<E> can be expressed, <E> =! E(x )P (x )dx =! En(E)P (E)dE where n(e) is the density of states and P (E) is the oltzmann 47 probability of a configuration with E.

48 he fact that s(e) is so small means that the contribution to <E > comes from a very narrow range of energies around a typical energy E*() that depends on the temperature n(e) P herefore, the partition function can be approximated by taking into account only the contribution related to E*(). E* Z =! n(e)exp[- E/k ] f (E*) = n(e*)exp[- E*/k ] \ F= k E*()+k ln n(e*) = E*()-S 48

49 he entropy, -k ln n(e*) is the logarithm of the degeneracy of the most probable energy. For a discrete system Z = Ê n(e i ) exp [- E i /k ] where E i are the set of energies of the system and n(e i ) their degeneracies. For a macroscopic system the number of different energies is large (~ for a discrete system) while only the maximal term n(e*) exp [- E*/k ] contributes. his product consists of two exponentials. At very low the product is maximal for the ground state energy, where most of the contribution comes from exp[- E GS */k ] while n(e GS *) ~ 1 (S=0). At very high the product is maximal for a high energy, where n(e*) is maximal (maximum degeneracy fl maximum entropy) but the exponential of the energy is small. For intermediate 49

50 n(e)e -E/k f (E) e -E*/k n(e) E*() E 50

51 he fact that the contribution to the integrals comes from an extremely narrow region of energies makes it very difficult to estimate <E>, S and other quantities by numerical integration. his is because the 3 dimensional configurational space is huge and the desired small region that contributes to the integrals is unknown a-priori. herefore, dividing the space into small regions (grid) would be impractical and even if done the corresponding integration would contribute zero because the small important region would be missed - clearly a waste of computer time. he success of Monte Carlo methods lies in their ability to find the contributing region very efficiently leading to precise estimation of various averages, such as <E>. 51

52 umerical integration f(x) x 1 x x 3 x n x! Ê i f(x i )Dx i Dx i = x i -x i-1 5

53 53 What is the probability to find the system in a certain energy (not x )? P (E)=n(E)exp-[E/k ]/Z So, this probability depends not only on the energy but also on the degeneracy n(e). he relative population of two energies is therefore: E E E k E E n E n k E k E E n E n E P E P = Δ Δ = = ) / exp( ) ( ) ( ) / exp( ) / exp( ) ( ) ( ) ( ) (

54 Solving problems in statistical mechanics he first step is to identify the states of the system and the corresponding energies (e.g., the configurations of a fluid and their energies). hen, three options are available: 1) he thermodynamic approach: Calculate the partition function Z fl the free energy F=-k lnz and obtain the properties of interest as suitable derivatives of F. ) Calculate statistical averages of the properties of interest. 3) Calculate the most probable term of Z and the most dominant contributions of the other properties. 54

55 Problem: independent spins interact with a magnetic field H. the interaction energy (potential) of a spin is mh or - mh, depending of whether m, the magnetic moment is positive or negative. Positive m leads to energy - mh. Calculate the various thermodynamic functions (E,F,S, etc.) at a given temperature. stats of the system because each spin is or +1(-m) or -1(m). Potential energy of spin configuration i: E i = - + mh + - mh or E i = -(- - )mh + - m H where and + - are the numbers of +1 and 1 spins. he magnetization of i is: M = + mh - - mh o kinetic energy is defined for this model. 55

56 Option 1: hermodynamic approach We have to calculate Z = Ê i exp-[e i /k ] ; i runs over all the different states of the system! his summation can be calculated by a trick. he spins are independent, i.e., they do not interact with each other fl changing a spin does not affect the other spins. herefore, the summation over the states of spins can be expressed as the product Z=(z 1 ) where z 1 is the partition function of a single spin (see oscillator!). z 1 = exp(-mh/k ) + exp(+mh/k )=cosh[mh/k ] cosh(x)=[exp(x)+exp(-x)]/ Z= [cosh(mh/k )] 56

57 F=-k lnz =-k ln[cosh(mh/k )] Entropy: F S = H = k μh ln{ cosh( k )} μh k μh tanh( k ) =, S/=ln ; =0, S=0 e x e x sinh( x) tanh( x ) = = e x + e x cosh( x) Energy: E = F + S = F H μh = μh tanh( ) k 57

58 58 Magnetization: M= ) tanh( k H H F M μ μ = = Specific heat: ) ( cosh k H k H k E C H μ μ = =

59 Option : Statistical approach Again we can treat first a single spin and calculate its average energy. z 1 for a single spin is: z 1 = exp(-mh/k ) + exp(+mh/k )= cosh[mh/k ] he oltzmann probability for spin is: exp[ mh/k ] / z 1 he average energy is <E> 1 = [-mhexp(+mh/k ) + mhexp(-mh/k )]/z 1 = -mh {sinh[mh/k ]}/ {cosh[mh/k ]}= = -mh tanh(mh/k ) ; <E>= -mhtanh(mh/k ) 59

60 he entropy s 1 of a single spin is: s 1 = -k [P + lnp + +P - lnp - ], where P is the oltzmann probability. s 1 = -k {exp[+mh/k ][mh/k -ln z 1 ] + exp[-mh/k ][-mh/k -ln z 1 ]}/z 1 = =-k {mh/k [e + -e - ]/ z 1 - ln z 1 [e + +e - ]/ z 1 }= = -k {mh/k tanh[mh/k ] ln z 1 } he same result as on p

61 Option 3: Using the most probable term E =- + m H + - m H fl E = E/mH = = (+ E )/ ; M = m ( ) = = (- E )/ E =-MH he # of spin configurations with E, W(E)=!/( +! -!) he terms of the partition function have the form, f! ( E ) = exp[ E μh / k] + E E!! 61

62 6 For a given,, and H we seek to find the maximal term. We take ln f (E), derive it with respect to E, and equate the result to 0, using the Stirling formula, ln! º ln. k H E E k H E E μ = + μ = + 1 exp ln he maximum or minimum of a function f(x) with respect to x, is obtained at the value x* where (df/dx) x* = f (x*)= 0 and (df /dx) x* <0 or >0, respectively. k HE E E E E E f μ + + = ln ln ln ) ( ln 0 ln 1 ln 1 ) ( = μ + + = k H E E E E f

63 63 ] tanh[ ) exp( ) exp( ) exp( ) exp( ) exp( ) exp( ) exp( ) exp( k H E k H k H k H k H k H k H k H k H E μ = μ + μ μ μ μ μ = μ + μ = 1 1 he most probable energy, E* for given and H is: E*=-mH tanh(mh/k ) and M= m tanh(mh/k ) As obtained before.

64 degeneracy energy typical spin configurations 1 (min. S=0) -mh (min.) 0 = 0 ÆÆÆÆ ÆÆ Æ(H) -(-1)mH+mH very low ( 0 ) ÆÆÆ ÆÆ = -mh+mh!/[(-)!!] =(-1)/ -mh+4mh > 0 ÆÆ ÆÆ. k!/[(-k)!k!] -mh +kmh k > k-1 ÆÆ.. / /! / [(/)!(/)!] mh (/-/) high..ææ ÆÆ (max. degeneracy =0 S = ln ) - > + not physical, negative temperature! de increases, ds decreases. 64

65 65

66 Several points: he entropy can be defined in two ways: 1) As a statistical average: S = -k Ê i P i lnp i (P i oltzmann) and ) as: k ln n(e*) n(e*) - degeneracy of the most probable energy. For large systems the two definitions are identical. As a mechanical system the spins would like to stay in the ground state (all spins are up; lowest potential energy fl most stable state), where no uncertainty exists (maximum order) fl the entropy is 0. 66

67 However, the spins interact with a heat bath at a finite, where random energy flows in and out the spin system. hus, spins parallel to H (spin up) might absorb energy and jump to their higher energy level (spin down), then some of them will release their energy back to the bath by returning to the lower energy state (Æ) and vice versa. For a given the average number of excited spins ( ) is constant and this number increases (i.e., the average energy increases) as is increased. he statistical mechanics treatment of this model describes this physical picture. As increases the average energy (= E*() - the most probable energy) increases correspondingly. 67

68 As E*() increases the number of states n(e*) with energy E*() increases as well, i.e., the system can populate more states with the same probability fl the uncertainty about its location increases fl S increases. So, the increased energy and its randomness provided by the heat bath as increases, is expressed in the spin system by higher E*() and enhanced disorder, i.e., larger ln n(e*) fl larger S(). he stability of a thermodynamic system is a compromise between two opposing tendencies: to be in the lowest potential energy and to be in a maximal disorder. At =0 the potential energy wins; it is minimal fl complete order S=0 (minimal). At = the disorder wins, S and E* are both maximal. 68

69 At finite the stability becomes a compromise between the tendencies for order and disorder; it is determined by finding the most probable (maximal) term of the partition function [at E*()]: n(e*) exp-[e*/k ] or equivalently the minimal term of the free energy, E* + k ln n(e*) = E*-S* otice that while the (macroscopic) energy is known very accurately due to the small fluctuations, the configuration (state) is unknown. We only know that the system can be located with equal probability in any of the n(e*) states. 69